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5-letter strings as DIGIT or DGIIT

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5-letter strings as DIGIT or DGIIT

by rsarashi » Sun Apr 16, 2017 5:32 am
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48

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by GMATGuruNY » Sun Apr 16, 2017 6:07 am
The letters D, G, I, I, and T can be used to form 5-letter strings such as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

(A) 12
(B) 18
(C) 24
(D) 36
(E) 48
Good arrangements = total arrangements - bad arrangements.

Total arrangements:
Number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical I's:
5!/2! = 60.

Bad arrangements:
In a bad arrangement, the two I's are in adjacent slots.
Let [II] represent the 2 adjacent I's.
Number of ways to arrange the 4 elements [II], D, G and T = 4! = 24.

Good arrangements:
Total arrangements - bad arrangements = 60-24 = 36.

The correct answer is D.
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by Brent@GMATPrepNow » Sun Apr 16, 2017 6:40 am
rsarashi wrote:The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48
Another approach:

Take the task of arranging the 5 letters and break it into stages.

Stage 1: Arrange the 3 CONSONANTS (D, G and T) in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 3 consonants in 3! ways (= 6 ways)
So, we can complete stage 1 in 6 ways

IMPORTANT: For each arrangement of 3 consonants, there are 4 places where the two I's can be placed. For example, in the arrangement DTG, we can add spaces as follows _D_T_G_
So, if we place each I in one of the available spaces, we can ENSURE that the two I's are never together.

Stage 2: Select two available spaces and place an I in each space.
Since the order in which we select the two spaces does not matter, we can use combinations.
We can select 2 spaces from 4 spaces in 4C2 ways (= 6 ways)
So we can complete stage 2 in 6 ways.

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 5 letters) in (6)(6) ways (= 36 ways)

Answer: D

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by [email protected] » Sun Apr 16, 2017 11:25 am
Hi rsarashi,

Since the two "I"s cannot be side-by-side, there are a limited number of ways to arrange the 5 letters. As such, with a little permutation math and some 'brute force', we can map out the possibilities:

_ _ _ _ _

If the first letter is an I, then the second letter CANNOT be an I (it would have to be one of the other 3 non-I letters)...

i 3

From here, any of the remaining letters can be in the 3rd spot. After placing one, either of the remaining two letters can be in the 4th spot and the last letter would be in the 5th spot...

i 3 3 2 1

This would give us (3)(3)(2)(1) = 18 possible arrangements with an I in the 1st spot.

If a non-I is in the 1st spot and an I is in the 2nd spot, then we have...

3 i _ _ _

A non-I would have to be in the 3rd spot, then either remaining letter could be 4th...

3 i 2 2 1

This would give us (3)(2)(2)(1) = 12 possible arrangements

Next, we could have two non-Is to start off, then Is in the 3rd and 5th spots...

3 2 i 1 i

This would give us (3)(2)(1) = 6 possible arrangements

There are no other options to account for, so we have 18+12+6 total arrangements.

Final Answer: D

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by Jay@ManhattanReview » Sun Apr 16, 2017 10:31 pm
rsarashi wrote:The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48

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Hi rsarashi,

What we want: 1. There is one letter among D, G, and T between the two 'I's. 2. There are two letters among D, G, and T between the two 'I's. 3. There are all three letter D, G, and T between the two 'I's.

What we do not want: No two 'I's are together.

Since it's easier to deal with 'What we do not want' than 'What we want,' let's go the other way.

# of ways the 'I's are separated = Total # of ways letters D, G, I, I, and T can make words (without any constraints) - Total # of ways the two 'I's are together


=> Total # of ways letters D, G, I, I, and T can make words (without any constraints) = 5! / 2! = 60; There are 5 letters (D, G, I, I, and T ) and two letters (I) are common.

=> Total # of ways the two 'I's are together = 4! = 24; considering the two 'I's as one letter

=> # of ways no 'I's are together = 60 - 24 = 36.

The correct answer: D

Hope this helps!

Relevant book: Manhattan Review GMAT Combinatorics and Probability Guide

-Jay
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by Jeff@TargetTestPrep » Mon Apr 24, 2017 4:06 pm
rsarashi wrote:The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48
This is a permutation problem, because the order of the letters matters. Let's first determine the total number of ways we can arrange the letters. Using the indistinguishable permutations formula, which takes into account the 2 repeating Is, we can arrange the letters in 5!/2! = 120/2 = 60 ways.

We also have the following equation:

60 = (number of ways to arrange the letters with the Is together) + (number of ways without the Is together).

Let's determine the number of ways to arrange the letters with the Is together.

We have: [I-I] [D] [G] [T]

We see that with the Is together, we have 4! = 24 ways to arrange the letters.

Thus, the number of ways to arrange the letters without the Is together (i.e., with the Is separated) is 60 - 24 = 36.

Answer: D

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